The reasons why FF sensors typically produce less noisy images in practice than APS-C sensors are that a) the (usually) bigger individual pixels produce a better signal-to-noise ratio, and b) the native image requires less enlargement when printing.

So you agree that a larger enlargement makes noise more visible? If so, it seems like we're agreeing on this, but arguing about semantics.

From my point of view, amplification and magnification amount to one and the same thing. Take two different sized sensors with the same number of MP (1D X and 7D for example), then the smaller sensor clearly has smaller pixels. Each pixel receives less light, so in order to give the same electrical signal to create a calibrated ISO 100, it has to amplify to a greater level the smaller signal created from the smaller number of photons collected at its smaller pixel. However, take two sensors of different size with pixels of the same density (eg 1Ds mk III and 30D, or D800 and D7000), and on a pixel level, each pixel is the same size, so it collects the same number of photons, creates the same strength electrical signal and requires the same amount of amplification. But it is magnified less from that larger sensor (each pixel, together with its noise is a less significant part of the whole image), so it's noise has a lower effect on the whole image. Not all FF and crop comparisons fall into one of these two convenient categories, so it's usually a combination of amplification and magnification differences between the two. But one thing is for sure - take the whole image, and the bigger sensor will be less noisy. And all things being equal (same generation technology etc), you'll find the noise ratio is directly in line with the area ratio.

Similarly speaking, a 1.6x crop of FF is no different from a 1.6x crop sensor. The exposure doesn't differ with different sized sensors. The noise does. Bigger area to capture light equals more light captured. That equals lower noise. Or you can turn it around by bumping up the ISO and create the same noise with greater sensitivity. That is what makes an f2.8 lens on FF equal to an f1.75 lens on 1.6x crop.

The reasons why FF sensors typically produce less noisy images in practice than APS-C sensors are that a) the (usually) bigger individual pixels produce a better signal-to-noise ratio, and b) the native image requires less enlargement when printing.

So you agree that a larger enlargement makes noise more visible? If so, it seems like we're agreeing on this, but arguing about semantics.

From my point of view, amplification and magnification amount to one and the same thing. Take two different sized sensors with the same number of MP (1D X and 7D for example), then the smaller sensor clearly has smaller pixels. Each pixel receives less light, so in order to give the same electrical signal to create a calibrated ISO 100, it has to amplify to a greater level the smaller signal created from the smaller number of photons collected at its smaller pixel. However, take two sensors of different size with pixels of the same density (eg 1Ds mk III and 30D, or D800 and D7000), and on a pixel level, each pixel is the same size, so it collects the same number of photons, creates the same strength electrical signal and requires the same amount of amplification. But it is magnified less from that larger sensor (each pixel, together with its noise is a less significant part of the whole image), so it's noise has a lower effect on the whole image. Not all FF and crop comparisons fall into one of these two convenient categories, so it's usually a combination of amplification and magnification differences between the two. But one thing is for sure - take the whole image, and the bigger sensor will be less noisy. And all things being equal (same generation technology etc), you'll find the noise ratio is directly in line with the area ratio.

Similarly speaking, a 1.6x crop of FF is no different from a 1.6x crop sensor. The exposure doesn't differ with different sized sensors. The noise does. Bigger area to capture light equals more light captured. That equals lower noise. Or you can turn it around by bumping up the ISO and create the same noise with greater sensitivity. That is what makes an f2.8 lens on FF equal to an f1.75 lens on 1.6x crop.

You were previously arguing that noise is directly related to the area of the sensor itself (i.e. APS-C needs 2.56x more signal amplification than FF) which is, of course, nonsense.

A FF sensor behind an f2.8 FF lens gathers 2.56x as much light as an f2.8 lens does on a 1.6x crop sensor due to the sensors 2.56x bigger surface area. If you only capture a fraction of all that FF f2.8 light by cropping it, well, the obvious happens from the light gathering point of view. The reason why using an f1.8 lens wide open on crop gives a brighter image than f2.8 on FF (when both are at the same ISO and shutter speed) is the amplification of the crop cameras sensor is 2.56x greater, at the expense of noise at any given ISO rating.

...to compare the 18-35/1.8 to the 24-70/2.8 on FF we'd need to set them both to a focal length to give an equal field of view. So for the sake of this example, lets use the long end of the Sigma's zoom - set the 18-35/1.8 to 35mm, which is equivalent of 56mm on the 24-70. We get the following:

Or think about it like this - imagine a photo taken with a FF lens and a FF sensor. Now you take that same photo and you crop out just the centre 40% - you've taken away 60% of the image - which is also 60% of the light that passed through that FF lens. You're left with only 40% of the light. That's what crop does. You need a faster lens on crop to make it capture the same amount of light in that smaller area.

A 15-35mm f/1.8 lens on crop, will deliver identical optical geometry to a 29-56mm f/2.88lens on full frame.

Depth of field for example, just like angle of view is a function of geometry.

So as the iris or aperture opens up more it allows more light in, but because that light is light that is coming at a higher incident angle it is focused over a greater area if it is behind or infront of the focal plane. So therefore you have a shallower depth of field. See how the red point is larger and therefore will be blurred on the open aperture example?

Here's another example:

People have been misled with crop to full frame conversions for years.

The "35mm equivalent" is what is really important and nothing else. Your images on 35mm equivalent will always look the same no matter what.

From a physics perspective the "35mm equivalent" is capturing identical information. What really matters is the geometry of the light hitting the sensor, and with 35mm equivalent the geometry will always be the same for a given equivalence. Not only that but your flash settings etc will be identical:

Going back to the 35mm equivalent discussion, consider this:

On APS-C (like the 7D) compared to full frame (like the 5D Mark III)

The sensor is 1.6 x 1.6 times smaller.

35mm equivalent aperture - Multiply F-Number by (1.6 ) . (an f stop is a base 2 log, so even though we have 1.6x1.6 times as much light we take the square root, which is 1.6 to multiply the F number by. (example 2.8 x 1.6 = 4.48, 4.0 x 1.6 = 6.4, 1.8 x 1.6 = 2.88))

35mm equivalent focal length - Multiply by 1.6

35mm equivalent ISO or light sensitivity - Multiply by (1.6 x 1.6) (bet you haven't heard of that, but if you do the math the an APS-C sensor amplifies the signal 1.6x1.6 times more at a given PIXEL than the a full frame camera, so even if both say ISO 800, ISO 800 on the an APS-C it is multiplying the light from each individual pixel the same as ISO 2000 full frame, assuming they had identical resolution.

This is an important point of contention. So while for a given area of a light sensetive material, ISO provides a set level of amplification, cameras do not have a consistent area that they absorb incoming light on. Some have finger nail sized sesnors, and some postage stamp sized sensors. What's more important to consider is each individual pixel and it's level of amplification. Images are formed from pixels, so to acheive a true equivalency between different sized sensors and produce identical images we must consider the data collected at each individual pixel.

Simply put APS-C cameras have more dense sensors with more pixels, even with identical resolution. Because ISO is dependent on a given volume of light passing through a given area, if we increase the number of pixels in that area each pixel will have a stronger amplification level.

To illustrate this imagine a rain storm. On one part of the ground you have a set of square buckets that are 1" x 1" x 10", on another you have a set of buckets that are 2" x 2" x 10". After a given identical volume of rain, the 2x2x10 buckets will have 4 times more liquid in them than the 1x1x10 buckets. However four 1x1x10 buckets will have the same amount of rain in them as a single 2x2x10 bucket, as they will cover the same area (4 square inches).

Now lets imagine you have to create a given level of brightness from the amount of liquid in each bucket. Because the smaller buckets have a smaller cross section you have to multiply the amount of liquid in them for each smaller bucket to produce the same level of brightness as a larger bucket.

So on the 2x2x10 bucket, 20 cubic inches of rain might equal a luminance value of 128. For the same amount of rain fall each 1x1x10 bucket would only contain 5 cubic inches of rain, but if you wanted to have the same luminance value you would have to multiply the volume of light at each pixel by 4.

In this way, a sensor with denser pixels always has to multiply the volume of light to get a given luminance value more than a sensor that is less dense.

ISO is a function of the volume of light resulting in a certain exposure ie the amount of rain coming from the sky, resulting in a given exposure level. Say 5 inches of rain per hour resulting in a luminance value of 255, and 2.5 inches resulting in 128 in our example.

So to collect identical information ie number of photons at each individual pixel, with an APS-C and FF camera that have identical resolutions, we must use a different volumes of light hitting the sensor, specifically on APS-C we must have a higher volume of light, which results in a lower ISO to produce the same level of exposure. Just like we would have to use a greater flow of rain to collect the same amount of water in a smaller bucket compared to a bigger bucket.

Because light can simply be compressed and expanded at will by changing the beam path, which alters the volume of light in a given area (think magnifying glass), all that we have to do when converting an incoming image from FF to APS-C is focus the same image coming through an identical iris on a smaller area to capture the same exact information on an APS-C sensor as a full frame sensor. If you go through the geometry and the math of doing this equivalency something amazing happens though. At "35mm equivalent" ISO, focal length and aperture and the same shutter speed, on a full frame and APS-C sensor if you have a theoretically perfect body and lens system you would get the exact same photons which came from the exact same sources from your subject landing in the same number and location in each individual coresponding pixel site and resulting in the exact same luminance values in the output for both cameras. With each camera having a different ISO setting, focal length, and aperture.

Identical photons from identical sources, landing in coresponding pixels, resulting in identical luminance values are what is important when desiring to create identical images. That is how 35mm equivalence works, and why it is important.

Discussing anything other than 35mm equivalent values for cameras is like saying I have a million dollars, and then failing to mention these are Zimbabwe dollars worth $20 not, American dollars.

Yes aperture ISO and focal length are fixed numbers, but so are monetary figures, and the most important thing even the most basic dealing of currency has is WHAT currency you're dealing with, and 99% of people require an "equivalent" frame of refference to understand foreign currency or need to do a conversion. Likewise with cameras, geometry (type of currency) is the most important thing when dealing with the performance of a camera system, and the first thing anyone needs to do is bring up a conversion to the local frame of reference, APS-C ,35mm, whatever.

Now of course each lens will have it's own characteristics and each body will likewise have it's own, and the full frame bodies and lenses tend to produce higher quality images due to the fact that miaturizing the system has adverse side effects, but if both bodies and lenses were theoretically perfect and had the same resolution these settings would deliver the exact same images with completely identical pixels.

35mm equivalent aperture - Multiply F-Number by (1.6 ) . (an f stop is a base 2 log, so even though we have 1.6x1.6 times as much light we take the square root, which is 1.6 to multiply the F number by. (example 2.8 x 1.6 = 4.48, 4.0 x 1.6 = 6.4, 1.8 x 1.6 = 2.88))

You'll have to clarify what you mean by "35mm equivalent aperture." The fact is, f/1.8 is f/1.8 whether mounted on FF or crop.

35mm equivalent ISO or light sensitivity - Multiply by (1.6 x 1.6) (bet you haven't heard of that, but if you do the math the an APS-C sensor amplifies the signal 1.6x1.6 times more at a given PIXEL than the a full frame camera, so even if both say ISO 800, ISO 800 on the an APS-C it is multiplying the light from each individual pixel the same as ISO 2000 full frame, assuming they had identical resolution, if resolution differs)

You're right, I haven't heard of that and the math eludes me. Please could you show it to us.

Edit: I reread your statement and saw that your calculation is "assuming they had identical resolution." So, in other words, the signal amplification of a sensor is related to pixel size, not sensor size. On that principle, we are agreed.

And on that same principle, the relationship of a lens's aperture to sensor size that has been made throughout this debate is spurious. It boils down to the same thing: small pixels are noisier than big pixels. If that is the only contention that this thread has been about, then we're all "in violent agreement."

Because the Sigma is an f/1.8 lens, it may be quite possible that the 24-70 does not create a bright enough exposure for a photographer in a low-light situation, so the Sigma will be much "nicer" to use in that situation.

What about using the FF system at a higher ISO? As the FF sensors larger area allows it to capture 2.56x more light, you can use an ISO 2.56x higher (just over a stop), without suffering from any more noise than the crop sensor. ISO 10,000 on a typical APS-C sensor gives the same noise as ISO 25,600 on a typical FF. If you do choose to make use of the higher ISO's made available to you, the final exposure is the same, and the f1.8 crop system offers no low light advantage over an f2.8 full frame system.

You're right, the newest FF sensors typically deliver the same ISO as a crop sensor could with less noise. But when the next generation of crop sensors come out, the difference won't be as big. And let's not forget that crop bodies cost significantly less than FF bodies. Not everyone has the budget for an FF camera.

Also, not everyone has the latest FF sensor. What if you're shooting in a low-light situation with a 5D classic (the original mark I version) with a 24-70 f/2.8 and a Rebel T5i with the new Sigma 18-35 f/1.8? I would personally use the Rebel+Sigma in that situation.

I haven't seen any RAW samples of the T5i (and I'm actually not too familiar with the 5D's files either) but I'd guess that the noise performance gap between those two cameras isn't that huge.

Let's say Canon releases the new 70D with a next-generation crop sensor, and costs $1200 at launch. (Likely to happen if Canon wants the camera to compete with the Nikon D7100. Additionally, I believe the price would drop shortly after release, just like the 6D)Now let's say this new sensor has noise performance as good as the 5D Mark II. (A stretch, but I think it's possible)And let's pretend that the Sigma 18-35 f/1.8 costs $1200. (Honestly, I think it will cost a little less than this)

So now you've got your Canon 70D with Sigma 18-35 f/1.8 (which you paid $2400 for) and you shoot it at ISO 100, f/1.8, 1/200s.Now you take your old 5D Mark II (which you bought used for $1200) and mount your 24-70 f/2.8 (the mark I version, which you also bought used for $1200) and shoot it at ISO 100, f/2.8, 1/200s.

Guess what? The crop image is going to be brighter than the FF image. And if the new, next-generation crop sensor really does have high-ISO noise performance as good as the 5D Mark II, then you also have a brand new crop setup that costs the same as a used FF setup and still gets a brighter exposure because of the large f/1.8 aperture than the f/2.8 on full frame, with the same noise at any ISO.

Now let's say this new sensor has noise performance as good as the 5D Mark II.

Not a chance.

Cheers,

b&

Take a look at the Fujifilm X-E1. I don't know what kind of tricks they're pulling but it certainly looks like they have the noise beaten down, smothered, crushed and trampled 'till it's unrecognisable as whatever it used to be.

While I agree with you, I don't agree that framing (and DoF) and exposure always have to be considered together. The crop format is mature enough to be considered on its own and without always being compared to its 35mm equivalent.

It is plain that f/1.8 on crop corresponds to around f/2.8 on FF in terms of DoF. The point of the discussion is whether or not a f/1.8 lens will give you the same exposure both on crop and FF.

You calculation seems to agree with that, since the distance from f/2.8 to f/4.5 and from ISO 800 to 2048 is always (more or less) 1.3 EV.